Question: Evaluate the iterated integral. $ \int_{-1}^2 \left( \int_0^2 x^3 e^y \, dx \right) dy =$ Choose 1 answer: Choose 1 answer: (Choice A) A $4(e^2 + e)$ (Choice B) B $4(e^2 - e^{-1})$ (Choice C) C $2(e^2 - e^{-1})$ (Choice D) D $e^{-1} - e^2$
Answer: Evaluate the inner integral: $\begin{aligned} \int_{-1}^2 \left( \int_0^2 x^3 e^y \, dx \right) dy &= \int_{-1}^2 \left[ \dfrac{x^4e^y}{4} \right]_0^2 dy \\ \\ &= \int_{-1}^2 4e^y \, dy \end{aligned}$ Evaluate the outer integral: $\begin{aligned} \int_{-1}^2 4e^y \, dy &= 4e^{y} \bigg|_{-1}^2 \\ \\ &= 4(e^2 - e^{-1}) \end{aligned}$ The answer: $ \int_{-1}^2 \left( \int_0^2 x^3 e^y \, dx \right) dy = 4(e^2 - e^{-1})$